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3 years ago

The chart shows the bids provided by four engineers to test a prototype.

Engineering

1 answer:

klasskru [66]3 years ago

8 0

Answer:

Explanation:

To know which is most or least cost-effective, it's not enough to look at only the per day rate, or only the time to complete. You have to multiply them to get the total cost of the project.

$\left[\begin{array}{ccccc}&Cost\ per\ day\ (\$)&Time\ to\ complete\ (days)&Total\ cost\ (\$)\\Zoe&500&8&4000\\Greg&650&10&6500\\Orion&400&12&4800\\Jin&700&5&3500\end{array}\right]$

As you can see, Greg is the least cost-effective because he charges the most for the project.

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A car accelerates from rest with an acceleration of 5 m/s^2. The acceleration decreases linearly with time to zero in 15 s, afte

Tpy6a [65]

Answer: At time 18.33 seconds it will have moved 500 meters.

Explanation:

Since the acceleration of the car is a linear function of time it can be written as a function of time as

$a(t)=5(1-\frac{t}{15})$

$a=\frac{d^{2}x}{dt^{2}}\\\\\therefore \frac{d^{2}x}{dt^{2}}=5(1-\frac{t}{15})$

Integrating both sides we get

$\int \frac{d^{2}x}{dt^{2}}dt=\int 5(1-\frac{t}{15})dt\\\\\frac{dx}{dt}=v=5t-\frac{5t^{2}}{30}+c$

Now since car starts from rest thus at time t = 0 ; v=0 thus c=0

again integrating with respect to time we get

$\int \frac{dx}{dt}dt=\int (5t-\frac{5t^{2}}{30})dt\\\\x(t)=\frac{5t^{2}}{2}-\frac{5t^{3}}{90}+D$

Now let us assume that car starts from origin thus D=0

thus in the first 15 seconds it covers a distance of

$x(15)=2.5\times 15^{2}-\farc{15^{3}}{18}=375m$

Thus the remaining 125 meters will be covered with a constant speed of

$v(15)=5\times 15-\frac{15^{2}}{6}=37.5m/s$

in time equalling $t_{2}=\frac{125}{37.5}=3.33seconds$

Thus the total time it requires equals 15+3.33 seconds

t=18.33 seconds

3 0

2 years ago

Airflow through a long, 0.15-m-square air conditioning duct maintains the outer duct surface temperature at 10°C. If the horizon

Ulleksa [173]

The complete Question is:

Airflow through a long, 0.15-m-square air conditioning duct maintains the outer duct surface temperature at 10°C. If the horizontal duct is uninsulated and exposed to air at 35°C in the crawlspace beneath a home, what is the heat gain per unit length of the duct? Evaluate the properties of air at 300 K. For the sides of the duct, use the more accurate Churchill and Chu correlations for laminar flow on vertical plates.

What is the Rayleigh number for free convection on the outer sides of the duct?

What is the free convection heat transfer coefficient on the outer sides of the duct, in W/m2·K?

What is the Rayleigh number for free convection on the top of the duct?

What is the free convection heat transfer coefficient on the top of the duct, in W/m2·K?

What is the free convection heat transfer coefficient on the bottom of the duct, in W/m2·K?

What is the total heat gain to the duct per unit length, in W/m?

Answers:

- 7709251 or 7.709 ×10⁶

- 4.87

- 965073

- 5.931 W/m² K

- 2.868 W/m² K